What is the molar solubility of silver carbonate ( Ag2CO3 ) in water? The solubility-product constant for Ag2CO3 is 8.1 × 10 -12 at 25°C.
The Correct Answer and Explanation is:
To find the molar solubility of silver carbonate (Ag2CO3) in water, we can use the solubility-product constant (Ksp) and set up an equilibrium expression.
The dissociation of silver carbonate in water is represented as: Ag2CO3(s)⇌2Ag+(aq)+CO32−(aq)\text{Ag}_2\text{CO}_3 (s) \rightleftharpoons 2 \text{Ag}^+ (aq) + \text{CO}_3^{2-} (aq)Ag2CO3(s)⇌2Ag+(aq)+CO32−(aq)
Let the molar solubility of Ag2CO3 be s. This means that, at equilibrium:
- The concentration of Ag+ will be 2s because 2 moles of Ag+ are produced for every mole of Ag2CO3 that dissolves.
- The concentration of CO3^2- will be s because 1 mole of CO3^2- is produced for every mole of Ag2CO3 that dissolves.
The solubility product constant (Ksp) expression is: Ksp=[Ag+]2[CO32−]K_{sp} = [\text{Ag}^+]^2 [\text{CO}_3^{2-}]Ksp=[Ag+]2[CO32−]
Substituting the concentrations at equilibrium: Ksp=(2s)2(s)=4s3K_{sp} = (2s)^2 (s) = 4s^3Ksp=(2s)2(s)=4s3
Now, we substitute the given Ksp value: 8.1×10−12=4s38.1 \times 10^{-12} = 4s^38.1×10−12=4s3
Solving for s: s3=8.1×10−124=2.025×10−12s^3 = \frac{8.1 \times 10^{-12}}{4} = 2.025 \times 10^{-12}s3=48.1×10−12=2.025×10−12 s=2.025×10−123≈1.27×10−4 mol/Ls = \sqrt[3]{2.025 \times 10^{-12}} \approx 1.27 \times 10^{-4} \, \text{mol/L}s=32.025×10−12≈1.27×10−4mol/L
Thus, the molar solubility of Ag2CO3 in water is approximately 1.27 × 10^-4 mol/L.
Explanation:
The process involves the dissociation of Ag2CO3 in water, where the solubility product constant Ksp describes the equilibrium between the solid salt and its ions in solution. By relating the concentrations of the ions to the solubility (s), we can solve for the molar solubility of Ag2CO3. The Ksp value is a crucial factor that determines the extent of dissociation, and from this, we can calculate the concentration of the dissolved salt at equilibrium.
